An unmarked police car traveling a constant 95km/h is passed by a speeder travel

Question

An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 115km/h .Precisely 2.50sec after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s/s, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Here is what I did

1. Find displacement in the 2.50 sec before police accelerates and while speeder is speeding away
Xp = 95 km/h -> 26.38 m/s times 2.50 sec = 65.972 m
Xs = 115 km/h -> 31.94 times 2.50 sec = 79.861 m

2. Afterwards, police accelerates but travels same displacement as the speeder

So I used xf = xi + vit + 1/2(at^2) where vi is initial velocity

Since speeder is not accelerating, his a is 0. I plug in for his xi, which is the xs from 1.

speeder: xf = 79.861 + 31.94t
police: xf = 65.972 + 26.38t + (1/2)(1.9)(t^2)

I set these equations equal to each other (xf = xf) and get 7.737 as my answer. Where did I go wrong? Please help, I did my best...

Explanation / Answer

final answer wll be t+2.50 s as t starts when they pass each other

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